决定JSON数据python解析平

Question

我已经有一段时间没有被困在一个我找不到想要的解决方案的主题上了。

示例：我有一个这样的 json：

{
    "SECTION": {
        "ID": 1,
        "COMMENT" : "foo bar ",
        "STRUCTURE" : {
            "LIEN" : [
                {
                    "from": "2020-01-01",
                    "to": "2020-01-03"
                },
                {
                    "from": "2020-01-04",
                    "to": "2999-01-07"
                }
            ]
        },
        "CONTEXTE":{
            "NATURE": {
                "text": "lorem smdlk fjq lsjdf mqjsh dflkq hs dfhkq g"
            }
        }

    }
}

我想要输出，例如：

{
    "SECTION.ID": 1,
    "SECTION.COMMENT": "foo bar ",
    "SECTION.STRUCTURE.LIEN.from": "2020-01-01",
    "SECTION.STRUCTURE.LIEN.to": "2020-01-03",
    "SECTION.CONTEXTE.NATURE.text": "lorem smdlk fjq lsjdf mqjsh dflkq hs dfhkq g"
}

{
    "SECTION.ID": 1,
    "SECTION.COMMENT": "foo bar ",
    "SECTION.STRUCTURE.LIEN.from": "2020-01-04",
    "SECTION.STRUCTURE.LIEN.to": "2999-01-07",
    "SECTION.CONTEXTE.NATURE.text": "lorem smdlk fjq lsjdf mqjsh dflkq hs dfhkq g"
}

有人知道我如何在 python 中执行此操作吗？非常感谢

Answer 1

听起来像是递归的经典案例；聚合路径，直到达到 "simple" 值，然后写入 "aggregated path"."key" 对："value"

Answer 2

我建议您使用 json Python 模块将 JSON 对象转换为 Python 对象。然后你可以使用递归。如果您使用 Python 3.5 或更高版本，以下代码可能是一个很好的起点：

import json

def flatten_helper(prefix, list_of_dict):
    res = []
    for i in list_of_dict:
        res_dict={}
        for k, v in i.items():
            res_dict['.'.join([prefix,k])]=v
        res.append(res_dict)
    return res

def flatten(x):
    if isinstance(x, list):
        res = []
        for ele in x:
            res = res + flatten(ele)
        return res
    else:
        res = [{}]
        for k, v in x.items():
            if (isinstance(v, dict) or isinstance(v,list)):
                new_res = []
                tempo = flatten(v)
                for r in res:
                    for t in tempo:
                        new_res.append({**r, **t})
                res = flatten_helper(k,new_res)
            else:
                for i, val in enumerate(res):
                    res[i][k]=v
        return res

jsonobj = '{"SECTION": {"ID": 1, "COMMENT" : "foo bar ", "STRUCTURE" : { "LIEN" : [{"from": "2020-01-01", "to": "2020-01-03"}, {"from": "2020-01-04", "to": "2999-01-07" }]}, "CONTEXTE":{"NATURE": {"text": "lorem smdlk fjq lsjdf mqjsh dflkq hs dfhkq g"}}}}'

pyobj = json.loads(jsonobj)
res = flatten(pyobj)

决定JSON数据python解析平

Decide JSON data python parsing flat

python

arrays

parsing

json

flat